How would you solve this equation?

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Answer 1 · 2018-05-02T04:50:03

#beta = 120^@, 240^@ " or " (2pi)/3, (4pi)/3#

#2cos^2 beta - 7 cos beta - 4 = 0#

#2 cos^2 beta + cos beta - 8 cos beta - 4 = 0#

#cos beta * (2 cos beta + 1) - 4 * (2cos beta + 1) = 0#

#(cos beta -4) * (2 cos beta + 1) = 0#

#cos beta = 4, -1/2#

Since cos can have a value between -1 & +1 only, 4 is not a valid solution.

#cos beta = -1/2#

Cosine is negative in II and III Quadrant.

#:. beta = 120^@, 240^@#

Answer 2 · 2018-05-02T05:00:58

#beta = 120^@, 240^@#.

This will be a piece of cake if you just assume a variable equal to #cos beta#.

So, Let's Assume #cos beta = x#.

Now the Equation is,

#2x^2 - 7x - 4 = 0#

#rArr 2x^2 -(8 - 1)x - 4 = 0# [Well, #7 = 8 -1#, right?]

#rArr 2x^2 - 8x + x - 4 = 0# [Distributive Property]

#rArr 2x(x - 4) + 1 (x - 4) = 0# [Grouping Like Terms]

#rArr (x - 4)(2x + 1) = 0# [Grouping Again.]

So, Either #x - 4 = 0 rArr x = 4#

Or, #2x - 1 = 0 rArr 2x = -1 rArr x = -1/2#

So, #cos beta = 4, -1/2#.

Now,

Cosine (#cos#) is a periodic function whose value ranges from #-1# to #1#. [#-1 <= cos x <= 1#]

So, #cos beta = -1/2#

Now, #beta = 120^@, 240^@#. [As #beta = [0^@, 360^@)#]

Hope this helps.